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Chapter 9: Problem 24

Find the radius of convergence of the given series. $$ \sum_{n=2}^{\infty}(-1)^{n+1} \frac{1 \cdot 3 \cdot 5 \cdots(2 n-3)}{2^{n} n !} x^{n} $$

Short Answer

Expert verified
The radius of convergence is 4.

Step by step solution

01

Identify the General Term

The series is given as \( \sum_{n=2}^{\infty} (-1)^{n+1} \frac{1 \cdot 3 \cdot 5 \cdots (2n-3)}{2^{n} n!} x^{n} \). The general term of the series, \( a_n \), is \( (-1)^{n+1} \frac{1 \cdot 3 \cdot 5 \cdots (2n-3)}{2^{n} n!} x^{n} \).
02

Simplify the General Term

To use the ratio test, simplify the expression \( 1 \cdot 3 \cdot 5 \cdots (2n-3) \). This product is the double factorial of \( (2n-3) \), which can be expressed recursively: \( (2n-3)!! = \frac{(2n-2)!}{2^{n-1} (n-1)!} \). Rewrite \( a_n \) as \( (-1)^{n+1} \frac{(2n-2)!}{2^{n-1} (n-1)! 2^{n} n!} x^n \), which simplifies to \( \frac{(2n-2)!}{2^{2n-1} n! (n-1)!} x^n \).
03

Apply the Ratio Test

To find the radius of convergence, use the ratio test: \( \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| \). Calculate \( a_{n+1} = \frac{(2n)!}{2^{2n+1} n! (n)!} x^{n+1} \). The ratio is \( \left| \frac{\frac{(2n)!}{2^{2n+1} n! (n)!} x^{n+1}}{\frac{(2n-2)!}{2^{2n-1} n! (n-1)!} x^n} \right| \).
04

Simplify the Ratio

Cancel common terms in the ratio: \( \frac{(2n)!}{2(n) \cdot (2n-1) \cdot (2n-2)!} \cdot \frac{1}{2^2 x} \). This simplifies further to \( \frac{(2n(2n-1))x}{4} \). As \( n \) approaches infinity, this simplifies to \( \frac{x}{4} \).
05

Determine the Radius of Convergence

For convergence using the ratio test, ensure \( \left| \frac{x}{4} \right| < 1 \), or equivalently, \( |x| < 4 \). So, the radius of convergence \( R \) is 4.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Ratio Test
The Ratio Test is a powerful method used to determine the convergence of an infinite series. It tells us how fast the terms of a series decrease. When applying the Ratio Test, you find the limit of the absolute value of the ratio of consecutive terms, \[ \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right|. \]
  • If the limit is less than 1, the series converges.
  • If the limit is greater than 1, the series diverges.
  • If the limit equals 1, the test is inconclusive.
In the context of finding the radius of convergence for a power series, the Ratio Test helps us identify the values of \( x \) for which the series converges. With the example series, applying the Ratio Test involves simplifying terms and evaluating the limit, ultimately leading to a condition \( |x| < 4 \), indicating a radius of convergence of 4.
Power Series
A power series is a series of the form \[ \sum_{n=0}^{\infty} a_n x^n, \] where \( a_n \) represents the coefficients of the series and \( x \) is a variable. Power series are useful because they can be used to represent functions in an "infinite polynomial" form.
  • Examples of power series include Taylor and Maclaurin series.
  • These series converge over some interval, the extent of which is described by the radius of convergence.
Power series are extensively used in calculus, particularly in solving differential equations and in analytic function representation. For the given series in the exercise, the variable \( x^n \) describes a power series with a radius of convergence that was calculated using the Ratio Test.
Double Factorial
The concept of a double factorial, denoted by \( n!! \), is an extension of the usual factorial operation. Instead of multiplying all positive integers up to \( n \), the double factorial multiplies every other integer starting from \( n \).
  • For odd \( n \), the double factorial is the product of all odd numbers up to \( n \).
  • For even \( n \), it includes all even numbers.
  • For example, \( 7!! = 7 \cdot 5 \cdot 3 \cdot 1 \) and \( 8!! = 8 \cdot 6 \cdot 4 \cdot 2 \).
In the given series, the double factorial is used for the product \( 1 \cdot 3 \cdot 5 \cdots (2n-3) \). It simplifies the expression and calculation of the terms involved, aiding in the simplification necessary for applying the Ratio Test and determining convergence.
Infinite Series
An infinite series is the sum of the terms of an infinite sequence. It may converge (approach a finite value) or diverge (grow without bound). Convergence is a key consideration, as only convergent series are useful in representing functions or defining numbers.
  • Finite series have a limited number of terms, while infinite series continue indefinitely.
  • The convergence of an infinite series depends largely on the behavior of its terms as \( n \to \infty \).
  • Key tests for convergence include the Ratio Test, Root Test, and integral tests, among others.
In mathematics, infinite series play a critical role in analysis and applied mathematics. For instance, the series in the exercise is infinite because it starts at \( n=2 \) and extends indefinitely. Convergence is determined using the Ratio Test to find the radius of convergence, showcasing if and where the series represents a valid function.

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Most popular questions from this chapter

Determine which series diverge, which converge conditionally, and which converge absolutely. $$ \sum_{n=1}^{\infty}(-1)^{n} \frac{n^{n}}{n !} $$

Find the Taylor series of the given function about \(a\). Use the series already obtained in the text or in previous exercises. $$ f(x)=\sin x ; a=\pi / 3 $$

Find the Taylor series of \(f\) about \(a\), and write out the first four terms of the series. $$ f(x)=\frac{1}{\sqrt{1-x^{2}}} ; a=0 $$

Determine whether the series converges or diverges. $$ \sum_{n=1}^{\infty}(-1)^{n} \cot \left(\frac{\pi}{2}-\frac{1}{n}\right) $$

Use the power series expansion for \(\left(e^{x}-1\right) / x\) to verify that $$ \lim _{x \rightarrow 0} \frac{e^{x}-1}{x}=1 $$
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